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Câu hỏi: Cho hàm số $y=f\left( x \right)$ liên tục trên $\left[ \dfrac{1}{3};3 \right]$ thỏa mãn $f\left( x \right)+x.f\left( \dfrac{1}{x} \right)={{x}^{3}}-x$. Giá trị tích phân $I=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}dx}=\dfrac{a}{b}$ với a, b là các số tự nhiên và phân số $\dfrac{a}{b}$ tối giản. Tính $S=2a+b$.
A. $S=17$.
B. $S=16$.
C. $S=25$.
D. $S=18$.
A. $S=17$.
B. $S=16$.
C. $S=25$.
D. $S=18$.
Ta có $I=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{{{x}^{3}}-x-x.f\left( \dfrac{1}{x} \right)}{{{x}^{2}}+x}dx}=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{{{x}^{2}}-1-f\left( \dfrac{1}{x} \right)}{x+1}dx}=\int\limits_{\dfrac{1}{3}}^{3}{\left( x-1 \right)dx}-\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( \dfrac{1}{x} \right)}{x+1}dx}$
Đặt $t=\dfrac{1}{x}\Leftrightarrow dt=-\dfrac{dx}{{{x}^{2}}}\Leftrightarrow dx=-\dfrac{dt}{{{t}^{2}}}$ và $\left\{ \begin{aligned}
& x=\dfrac{1}{3}\Rightarrow t=3 \\
& x=3\Rightarrow t=\dfrac{1}{3} \\
\end{aligned} \right.$
Do đó $\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( \dfrac{1}{x} \right)}{x+1}dx}=\int\limits_{3}^{\dfrac{1}{3}}{\dfrac{1}{t}.\dfrac{f\left( t \right)}{1+\dfrac{1}{t}}.\left( -\dfrac{1}{{{t}^{2}}} \right)dt}=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( t \right)}{{{t}^{2}}+t}dt}=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}dx}$
Vậy $I=\int\limits_{\dfrac{1}{3}}^{3}{\left( x-1 \right)dx}-I\Rightarrow 2I=\dfrac{16}{9}\Rightarrow I=\dfrac{8}{9}=\dfrac{a}{b}\Rightarrow S=25$.
Đặt $t=\dfrac{1}{x}\Leftrightarrow dt=-\dfrac{dx}{{{x}^{2}}}\Leftrightarrow dx=-\dfrac{dt}{{{t}^{2}}}$ và $\left\{ \begin{aligned}
& x=\dfrac{1}{3}\Rightarrow t=3 \\
& x=3\Rightarrow t=\dfrac{1}{3} \\
\end{aligned} \right.$
Do đó $\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( \dfrac{1}{x} \right)}{x+1}dx}=\int\limits_{3}^{\dfrac{1}{3}}{\dfrac{1}{t}.\dfrac{f\left( t \right)}{1+\dfrac{1}{t}}.\left( -\dfrac{1}{{{t}^{2}}} \right)dt}=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( t \right)}{{{t}^{2}}+t}dt}=\int\limits_{\dfrac{1}{3}}^{3}{\dfrac{f\left( x \right)}{{{x}^{2}}+x}dx}$
Vậy $I=\int\limits_{\dfrac{1}{3}}^{3}{\left( x-1 \right)dx}-I\Rightarrow 2I=\dfrac{16}{9}\Rightarrow I=\dfrac{8}{9}=\dfrac{a}{b}\Rightarrow S=25$.
Đáp án C.